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# The area of the region bounded by the y-axis ,$y=\cos x$ and $y=\sin x,0\leq x\leq \large\frac{\pi}{2}$ is\begin{array}{1 1}(A)\;1\sqrt 2sq.units & (B)\;(\sqrt 2+1)sq.units\\(C)\;(\sqrt 2-1)sq.units & (D)\;(2\sqrt 2-1)sq.units \end{array}

Toolbox:
• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\sin(A-B)=2\cos\large\frac{(A+B)}{2}$$\sin\large\frac{(A-B)}{2} • Principle solution of \sin\theta=n\pi\pm(-1)^n\alpha • \int \cos \theta=\sin\theta+c Step 1: We have to find the area of the region bounded by the y-axis,y=\cos x and y=\sin x. Where x ranges between 0\leq x\leq \large\frac{\pi}{2} First to find the area,let us find the point of intersection of f(x)=\sin x and g(x)=\cos x At the point of intersection \sin x=\cos x \Rightarrow \sin x=\sin(\large\frac{\pi}{2}$$-x$)=0
Now applying $\sin(A-B)=2\cos\large\frac{(A+B)}{2}$$\sin\large\frac{(A-B)}{2},we get \;\;=2\cos \big(\large\frac{x+\Large\frac{\pi}{2}-x}{2}\big)\normalsize\sin\big(\large\frac{x-(\Large\frac{\pi}{2}-x)}{2}\big)$$=0$
$\;\;=2\cos\large\frac{\pi}{4}$$\sin(x-\large\frac{\pi}{4})$$=0$
But $\cos\large\frac{\pi}{4}=\large\frac{1}{\sqrt 2}$
$\Rightarrow 2.\large\frac{1}{\sqrt 2}$$\sin(x-\large\frac{\pi}{4})$$=0$
$\sin(x-\large\frac{\pi}{4})$$=0\Rightarrow x-\large\frac{\pi}{4}$$=0$
Step 2:
Applying the principle solution we get,
$x-\large\frac{\pi}{4}=$$n\pi \Rightarrow x=\large\frac{\pi}{4},\frac{5\pi}{4},\frac{9\pi}{4}........ But it is given x values ranges between 0 and \large\frac{\pi}{2} Hence we take only \large\frac{\pi}{4} and discard the 0 points. Required area is the shaded portion shown in the fig. The area of the shaded portion is given by A=\int_0^{\Large\frac{\pi}{4}}(\sin x-\cos x)dx. Step 3: On integrating we get, A=\begin{bmatrix}-\cos x-\sin x\end{bmatrix}_0^{\Large\frac{\pi}{4}} On applying the limits we get, A=[-(\cos\large\frac{\pi}{4}-$$\cos 0)-(\sin\large\frac{\pi}{4}$$-\sin 0)] But \sin\large\frac{\pi}{4}=$$\cos\large\frac{\pi}{4}=\frac{1}{\sqrt 2},$$\cos 0=1 and \sin 0=0 Substituting the values we get, \;\;\;=[-[\large\frac{1}{\sqrt 2}$$-1]-[\large\frac{1}{\sqrt 2}]]$
$\;\;\;=1-\large\frac{2}{\sqrt 2}=$$(1-\sqrt 2)$ or $\sqrt 2-1$
$A=(\sqrt 2-1)$sq.units.
Hence C is the correct option.
edited May 6, 2013